Go beyond basic deduction. Learn the powerful techniques of Conditional Proof and Reductio ad Absurdum to solve complex arguments and test for validity with speed and elegance.
1. A More Powerful Tool for Proofs
While the 19 rules of inference and replacement from Module 8 are powerful, some valid arguments are incredibly difficult—or even impossible—to prove with them alone. This is especially true for arguments whose conclusions are conditional statements (e.g., of the form P ⊃ Q).
The Conditional Proof (CP) method provides a new, elegant strategy. Instead of working directly from premises to a conditional conclusion, we can make a temporary, strategic assumption.
The Core Idea of Conditional Proof: To prove a conditional statement like A ⊃ C, you temporarily assume the antecedent (A) is true. Then, using the standard rules of deduction, you derive the consequent (C). If you succeed, you have proven that if A were true, C would necessarily follow. You can then discharge your assumption and formally conclude A ⊃ C.
The justification for the temporary assumption is called the Conditional Proof Assumption, abbreviated as CPA or, more commonly, ACP.
2. The Mechanics of Conditional Proof
To keep track of our temporary assumptions, we use a special notation to clearly define their scope. The scope includes the assumption itself and all the lines derived from it. Once we discharge the assumption, we can no longer refer to any lines inside its scope.
We'll visualize this with a vertical scope line. Everything inside this line depends on the temporary assumption.
Example: Proving a Hypothetical Syllogism
Let's prove A ⊃ C from the premises A ⊃ B and B ⊃ C without using the H.S. rule itself.
At line 3, we assume A. We use it to derive C at line 5. Having achieved our goal, we end the scope, "discharge" the assumption, and conclude A ⊃ C on line 6, justifying it by the entire sub-proof (lines 3 through 5) and the rule of Conditional Proof (C.P.).
Nested Conditional Proofs
The power of CP becomes even more apparent when we nest assumptions. We can make a new assumption inside the scope of another one. However, we must discharge them in the reverse order they were made—you must close an inner scope before closing the outer scope it's in.
A Nested Proof
This complex proof requires assuming L, and then, within that scope, assuming M.
3. Reductio ad Absurdum (Indirect Proof)
Closely related to CP is another powerful technique: Reductio ad Absurdum ("reduction to absurdity"), also known as Indirect Proof (I.P.). This method has been used for centuries in logic and mathematics.
The Core Idea of Reductio ad Absurdum: To prove a statement P is true, you temporarily assume it is false (~P). You then use this assumption to derive an explicit contradiction (e.g., Q • ~Q). Since your assumption led to an absurdity, the assumption itself must be false, meaning the original statement P must be true.
This is the logical engine behind the shorter truth table technique, which provides a fast way to test an argument for validity.
4. Practical Application: The Shorter Truth Table
The shorter truth table technique is a quick and powerful way to test an argument's validity, building directly on the principles of truth tables we explored in Module 5. It's a practical application of the Reductio ad Absurdum method. Instead of building a full truth table, you try to find a single "counterexample" row—a scenario where all premises are true and the conclusion is false.
The strategy is to assume the argument is invalid and see if that assumption leads to a logical contradiction.
Shorter Truth Table Logic:
1. Assume the argument is invalid (True premises, False conclusion).
2. Work backwards to find the required truth values for the simple statements.
3. If you find a contradiction (e.g., a statement must be both T and F), the assumption was impossible. The argument is VALID.
4. If you find a consistent set of values, you've found a counterexample. The argument is INVALID.
Example 1: Proving a Valid Argument
Let's test A ⊃ B, ~B /∴ ~A using the shorter truth table method.
First, write it as a single conditional: [(A ⊃ B) • ~B] ⊃ ~A. Now, let's assume this can be FALSE.
- For the main
⊃to be F, the antecedent[(A ⊃ B) • ~B]must be T and the consequent~Amust be F. - If
~Ais F, then A must be T. - Now look at the antecedent. For
[(A ⊃ B) • ~B]to be T, both parts must be T.(A ⊃ B)must be T.~Bmust be T, which means B must be F.
- Finally, let's check for a contradiction. We have determined that
Ais T andBis F. Does this work with our requirement that(A ⊃ B)is T? - Let's substitute:
(T ⊃ F). But a conditional with a true antecedent and a false consequent is FALSE. - CONTRADICTION! We needed
(A ⊃ B)to be T, but the values we were forced to assign make it F. Our initial assumption was impossible. Therefore, the argument is VALID.
Example 2: Proving an Invalid Argument
Let's test: A ⊃ B, B /∴ A (This is the fallacy of Affirming the Consequent).
Statement form: [(A ⊃ B) • B] ⊃ A. Assume it's FALSE.
- For the main
⊃to be F,[(A ⊃ B) • B]must be T andAmust be F. - We now know that A must be F.
- For the antecedent
[(A ⊃ B) • B]to be T, both parts must be T.(A ⊃ B)must be T.Bmust be T.
- Let's check for a contradiction. We have determined that A is F and B is T.
- Substitute these values into
(A ⊃ B). This gives us(F ⊃ T), which is TRUE. - NO CONTRADICTION! We have found a consistent assignment of truth values (
A=F,B=T) that makes both premises true but the conclusion false. Therefore, the argument is INVALID.
5. Test Your Understanding
Apply these advanced strategies to analyze the following questions.
1. When starting a Conditional Proof to prove the conclusion R ⊃ S, what is your very first step inside the proof?
2. The method of Indirect Proof (RAA) aims to derive what from an assumption?
3. You use the shorter truth table method and discover it's impossible to assign values without a contradiction. What does this mean?
4. After you finish a sub-proof for CP and write the final conditional line (e.g., A ⊃ C), can you use the individual line C from inside the sub-proof later on?